Question

Question #9bee2

Answer 1

See explanation.

Explanation:

We want to factor the trinomial

`x^8+5x^4+4`

This may look intimidating, since it's an octic equation. However, this becomes much more simple and familiar if we set `u=x^4`, yielding the trinomial

`u^2+5u+4`

This can simply be factored into

`(u+4)(u+1)`

Since `u=x^4`, this is equivalent to

`(x^4+4)(x^4+1)`

We can continue by factoring these using imaginary numbers.

Both of these terms can be factored as differences of squares, which take the form

`a^2-b^2=(a+b)(a-b)`

You may be asking yourself, how can these be differences of squares when the terms are being added?

The remedy to which is that the expression also equals

`(x^4-(-4))(x^4-(-1))`

If we treat `x^4-(-4)` as `a^2-b^2`, we see that `a^2=x^4` and `b^2=-4`, so `a=sqrt(x^4)=x^2` and `b=sqrt(-4)=2i`.

Thus, the expression with a factored first term is

`(x^2+2i)(x^2-2i)(x^4-(-1))`

We use a very similar method to factor `x^4-(-1)`, namely `a^2=x^4` and `b^2=-1`, so `a=sqrt(x^4)=x^2` and `b=sqrt(-1)=i`, hence the expression equals

`(x^2+2i)(x^2-2i)(x^2+i)(x^2-i)`